By Vladithur, history, 13 months ago,

Thanks for participating in the round, we hope you liked the problems!

Solve count predictions (official div. 2)

1712A - Wonderful Permutation

Hint
Tutorial

Bonus: solve for every $k$ from $1$ to $n$ for $n \le 10^5$.

1712B - Woeful Permutation

Hints
Tutorial

Bonus: try to prove the solution without the editorial!

1712C - Sort Zero

Hints
Tutorial

Bonus: solve for when $a_i$ can also be negative.

1712D - Empty Graph

Hints
Tutorial

Bonus: solve for every $k$ from $1$ to $n$.

1712E2 - LCM Sum (hard version)

Hints
Tutorial

Bonus: solve the problem in $\mathcal{O}((n + t) \log n)$ or better.

1712F - Triameter

Hints
Tutorial

Bonus: solve for $n, q \le 10^6$.

Don't forget to rate the problems!

Problem Feedback

PS: Solution codes probably will be added later.

UPD: explanations of the references:

• +325

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 » 13 months ago, # |   +3 Nice :)
•  » » 13 months ago, # ^ |   0 Why did you say that? What's the point of making such remarks? Look, this sentence of yours wasted one precious second of my life, and the reply wasted 30 seconds.
•  » » » 4 months ago, # ^ |   0 And reading the comment is wasting y'r life
•  » » » » 3 months ago, # ^ |   0 true bro
 » 13 months ago, # |   0 ;(
 » 13 months ago, # |   0 the gap between C and D is too large.
•  » » 13 months ago, # ^ |   0 Maybe yes, but E1 was solvable, you could solve E1 instead of D
•  » » 13 months ago, # ^ |   +39 We expected C to be a bit harder and D to be a bit easier.
•  » » » 13 months ago, # ^ |   -33 If you know Graph algorithms I think solving D would be easy however in general E1 is much easier
•  » » » » 13 months ago, # ^ |   +12 maybe. But for me, I solve D in 30 min but solve E1 in 1h30min(failed to find 6 and 15's apperance, not familiar with the method). And I think D doesn't really have much to do with Graph Algorithms.
•  » » » » » 13 months ago, # ^ |   0 Can you please explain the 6 and 15 case, cuz i dont understand it at all?
•  » » 13 months ago, # ^ |   +3 D isn't so hard, just D, like the ancient times, the C was easy let's gonna admit that
 » 13 months ago, # |   +40 Problem D is one of the best problems I’ve ever seen. Overall, round is awesome. Hope to see more such great contests!
•  » » 6 weeks ago, # ^ |   0 Problem D is one of the best problems I’ve ever seen. Overall, round is awesome. Hope to see more such great contests!
 » 13 months ago, # | ← Rev. 2 →   +16 If the formal proof is quite hard, then why is it problem B ?Do you expect people to solve without proof ?
•  » » 13 months ago, # ^ |   +4 That's a mistake on my part, I started wondering about the formal proof only on the day of the round, and it turned out to be unexpectedly difficult.
•  » » 13 months ago, # ^ |   +56 Do you expect people to solve without proof ? That's what people usually do...
•  » » » 13 months ago, # ^ |   +4 Yes, but if you guess it wrong, you're gonna get huge amount of penalties and waste a lot of time.
•  » » » 13 months ago, # ^ |   +12 Yea, but for most easy problems usually the gap between intuition and formal proof is not this large. Most of the time all you need to do is to formalize your intuition to turn it into a complete proof, which is definitely not the case for this problem.
•  » » 13 months ago, # ^ | ← Rev. 2 →   +12 Do you really think that those 13062 participants got Accepted on B all with proofs? :)
•  » » » 13 months ago, # ^ |   +23 Of course not, it's bad that they solved it without proof. My point is, for easy questions, make the proofs easy too.
•  » » » » 13 months ago, # ^ |   0 Or Make the problem with easy proofs. >_< (XD)
•  » » 13 months ago, # ^ |   0 While I do think that it may not be appropriate for div2B when the proof is so difficult, I think it's quite intuitive that the optimal result arises when big numbers are paired with big numbers, leaving small numbers paired with small numbers (as opposed to pairing big numbers with small numbers). Even without that intuition, trying it on a few small examples should quickly demonstrate this trend. Add in the observation that adjacent numbers have a GCD of 1 (so none of the factors are "wasted" by the LCM) and we have a straightforward solution that's very easy to implement. I didn't prove it during the contest myself, and that did cause me to hesitate before submission, but the intuition was strong enough that I decided to go ahead anyway.
•  » » » 13 months ago, # ^ |   0 But the intuition was strong enough that I got +2 :(
•  » » » » 13 months ago, # ^ |   0 I think your intuition was correct, but you simply didn't test odd values. The 1 should have been at the beginning, not the end.
•  » » 13 months ago, # ^ |   +2 rearrangement inequality, and the fact: x and x+1 are coprimes, is a little bit hard to proof but the intuition helps. Sadly many Competitive programmers don't like maths
•  » » 13 months ago, # ^ |   +3 i just wrote brute force and saw the pattern for (1 to 9) ;) it was just swap(arr[i],arr[i-1]) from back....
 » 13 months ago, # |   0 Can E2 be solved using sqrt decomposition?
 » 13 months ago, # |   0 The editorial doesn’t show up in the bottom right of the problem pages
•  » » 13 months ago, # ^ |   0 Now it should
 » 13 months ago, # |   +81 I accepted f in the contest, my solution is $O(qn \log n)$. Maybe because of the large constant, it fst.
•  » » 13 months ago, # ^ |   +64 And I don't know why it get wa when I optimize it.
•  » » » 13 months ago, # ^ |   +50 for vertices i,j the distance shall be $min(f[i]+f[j]+x,dis(i,j))$, f[i] means the distance between i and the nearest leaf. I sort f, and check whether the answer can be bigger than ans. Then for every node, it is a suffix of j that $f[i]+f[j]+x>=ans$ . I prework the suffix diameter and use O(1) lca to find the distance.
 » 13 months ago, # |   +18 This was challenging and editorial is awesome. Thanks for these competitions
 » 13 months ago, # |   +6 C can also be solved using binary search.Find the minimum $k$ such that the array remains sorted if the first unique $k$ instances of the array are converted to $0$.Solution: 168108392
 » 13 months ago, # |   0 Easiest Div2C ever?
•  » » 13 months ago, # ^ |   0 Yes probably. Some newbies solved all A,B,C and still got negative delta.
 » 13 months ago, # | ← Rev. 2 →   0 Can someone please tell me why i get WA with this source for problem D (greedy solution)?
•  » » 13 months ago, # ^ |   0 Take a look at Ticket 16031 from CF Stress for a counter example.
•  » » » 13 months ago, # ^ |   0 can you check for C ; https://codeforces.com/problemset/submission/1712/170962914
•  » » 13 months ago, # ^ |   +1 mircea_007 I did the same using priority queue but end up getting WA.
 » 13 months ago, # |   +1 Alternate proof of BAfter the first two paragraphs of the editorial, another way is to proceed by induction. By Rearrangement Inequality, in any optional permutation, if $i  » 13 months ago, # | 0 For subproof in problem D: When Imin is < u, can we take u to Imin and then Imin to v path instead of going till node 1? cc: Vladithur •  » » 13 months ago, # ^ | 0 Yes, going to node 1 just ensures that we always cross the min value.  » 13 months ago, # | +8 I made video Solutions to the first 4 problems in case people are interested.  » 13 months ago, # | 0 I don’t think we need that proof to solve B I just wrote the numbers down and i got my answers right away But btw nice editorial < 3 •  » » 13 months ago, # ^ | +1 During the contest you may be right,but I don't think you can learn something from the problem after the contest.Actually,if it was D or E,probably you would hesitate to submit.  » 13 months ago, # | +6 How do you arrive at the conclusion: a tuple$i \lt j \lt k$is bad when$\mathrm{lcm}(i,j,k) = 2 \cdot k$and$i+j \gt k$? •  » » 13 months ago, # ^ | +18$lcm(i, j, k) = 2 \cdot k \implies 2 \cdot k < i + j + k \implies i + j > k$•  » » » 13 months ago, # ^ | +6 I meant to ask how you get$\mathrm{lcm}(i,j,k) = 2 \cdot k$in particular. Why$2 \cdot k$? •  » » » » 13 months ago, # ^ | +35$lcm(i, j, k) < 3 \cdot k$, and$lcm$must divide$i$,$j$, and$k$. This leaves us with only two possible values for it:$k$and$2 \cdot k$. •  » » » » » 13 months ago, # ^ | +1 Ohh okay. Thanks!  » 13 months ago, # | +5 as a #008000 rated, I enjoyed solving C  » 13 months ago, # | ← Rev. 2 → 0 Could someone point out my mistake in problem D? I don't understand what the tute is saying very well. https://codeforces.com/contest/1712/submission/168169178 I've also made this edit, but it still didn't work. https://codeforces.com/contest/1712/submission/168199569 •  » » 13 months ago, # ^ | +7 Greedily changing the$k$smallest values isn't always correct •  » » » 13 months ago, # ^ | 0 Thanks, I hadn't thought of that. •  » » » 13 months ago, # ^ | ← Rev. 2 → 0 I did this at first as well  » 13 months ago, # | 0 Difficulties like ->Easy -> Easy -> Easy -> Hard -> Hard -> Hard -> Hard •  » » 13 months ago, # ^ | -16 E1 and E2 not hard •  » » » 13 months ago, # ^ | +8 you didnt solve them tho?  » 13 months ago, # | +3 Where does the "/6" and "/15" come from at problem E1? I ve seen this on many solutions. I guess its an optimization for finding all triplets i,j,k with a fixed k, but i dont undersand it. •  » » 13 months ago, # ^ | +3 You will see, if(i==6 j==10 k==15), lcm(i,j,k)=30 < i+j+k, if(i==3 j==4 k==6), lcm(i,j,k)=12 < i+j+k, and you may enlarge i,j,k because if lcm(i,j,k) < i+j+k ,then X(lcm(i,j,k)) < X(i+j+k) ,since X is a positive integer.And why is the cases (i==6 j==10 k==15) and (i==3 j==4 k==6) ? That's because 1/5+1/3+1/2 > 1 and 1/4+1/3+1/2 > 1 •  » » 13 months ago, # ^ | +49 This is for finding$(i, j, k)$where$lcm(i,j,k)=2k$and$i+j>k$.Note that because of the second condition,$\frac{k}{2} < j < k$. Since$j$divides$2k$,$j$must be$\frac{2k}{3}$.Using the second condition again,$\frac{k}{3}
•  » » » 13 months ago, # ^ |   0 Nice. Thanks for the clear explanation.
 » 13 months ago, # |   0 Very easy greedy Solution of problem D
•  » » 13 months ago, # ^ |   0 can you explain it as well plz?
•  » » » 13 months ago, # ^ |   0 Try this for an explanation.As discussed elsewhere, when doing an update of an a_i you should always update to 1e9. There are two possible strategies for doing the K updates of the a_i: 1. Update the smallest K of the a_i, or 2. Update the smallest (K-1) of the a_i, and then use the last update to update a neighbour of the largest of the updated a_i.We can split the 2nd strategy into two: if K=1, then the final update is to the largest of the original a_i if K>1, then the final update will be to a neighbour of a_i already updated with value 1e9.The final diameter of the graph is min( 2*smallest a_i, max of smaller neighbours a_i, a_i+1 )So overall we have to calculate 4 numbers: v11 — strategy 1 — 2*smallest of updated a_i v12 — stratgey 1 — max of smaller of the a_i and a_i+1 v21 — strategy 2 — 2*smallest of updated a_i v22 — stratgey 2 — max of smaller of the updated a_i and a_i+1diameter for strategy 1 is min(v11,v12) diameter for strategy 2 is min(v21,v22)We pick the strategy for the larger diameter, so the final answer max( min(v11,v12), min(v21,v22) )In the code: v22 is cnt — initially calculated for K=1, and overwritten with 1e9 if K>1 v21 is expressed as 2ll*vec[k-1].ff min(v21,v22) is put into "ans"The smallest K values of a_i are identified by populating and sorting vector> vec. Then we explicitly update the a_i array.v12 is calculated in maxi2 in a loop that looks at the neighbours min(a[i], a[i+1]); v11 is picked up from 2ll*vec[0].ff and we have maxi2 = min( maxi2, 2ll*vec[0].ff ) to calculate min(v11,v12).The final output cout << max( ans, maxi2) selects the diameter from the better of the two strategies.
•  » » » 13 months ago, # ^ |   0
•  » » » » 13 months ago, # ^ |   +3 thanks for the vid. much appreciated!
 » 13 months ago, # |   +1 Finally became CM! Like B & D very much.
 » 13 months ago, # |   0 Why should i to find 6 and 15 in problem E1? How can I think that way. I see tourist's solution seemed iterate in 30 to find such special case. It shows that he is trying to find such special case.
•  » » 13 months ago, # ^ |   0 For small numbers multiplication funtion has a low growing, so the idea is use BruteForce for those small values
 » 13 months ago, # | ← Rev. 2 →   +8 Did anybody solve this clue by the author? I'm still stuck on it.Edit: I saw the answer in editorial.
 » 13 months ago, # |   +4 I really think this round is quite exquisite. The bonus in every problem, the riddle in the beginning of each task(and ACGN lover here too!), and the nice problems.Though I was one of those who was stuck in problem D just in lack of a binary search while using a bad greedy, I still have to upvote the problem preparation for all involvers！Thankyou！
 » 13 months ago, # |   0 B Bonus: try to prove the solution without the editorial!Me: Use the data!
 » 13 months ago, # | ← Rev. 2 →   0 Could anyone please explain for what mysterious reasons this solution for D using ordered_set is not passing even small test cases like (n <= 1000)? (168216695)Hand written Treap passed pretests, but FST'ed mostly likely due to bug in my implementation (168172650)
 » 13 months ago, # |   +4 In my opinion, E1 and E2 are good problems. Thanks for the great contest!
 » 13 months ago, # |   0 In C iterating from last index, I checked if Ai < Ai-1. If yes then simply print number of distinct elements before Ai and break. If no then check whether count of Ai>1 && Ai!=Ai-1(i.e its duplicate lies in prefix of the array) then also print number of distinct elements before Ai and break.https://codeforces.com/contest/1712/submission/168175298Please help me where am I doing wrong
•  » » 13 months ago, # ^ |   0 I am not confident that your idea would work in general, but here is an immediate error I found: when line 34 is executed, i.e., decreasing the hash value of an element if it appears multiple times in a row, the decrement --c at line 37 is still executed, even though the number of distinct values in the prefix has not decreased. Here is are three test cases I encourage you to check: 3 4 1 3 2 2 5 1 3 2 2 2 6 1 3 2 2 2 2 Each occurrence of 2 decreases the number of distinct characters, with the final test case even reporting a result of -1. I think you intended to put a continue; after line 34. I'm still a little skeptical of whether this would be enough to correct the program (since the logic feels like it still has loopholes), but that should be a good step forward.
 » 13 months ago, # |   0 For prob D can someone explain this TC? 6 2 2 1 2 4 2 1 My solution is to replace the 2nd and 6th values with $10^9$. So the diameter here would be $d(2,6) = 4$ (going through one of the nodes with value 2), but the expected result is 2. Am I missing something here?
•  » » 13 months ago, # ^ |   +3 The length of the edge connecting node $2, 6$ is $2$, because $\min{a_2, a_3, a_4, a_5, a_6} = 2$.
•  » » » 13 months ago, # ^ |   0 Oh shit I misunderstood the problem for the whole night. Thank you!
 » 13 months ago, # | ← Rev. 4 →   +3 In fact, we can prove in E1, $j = \frac{2}{3}k$ if $\operatorname{lcm}(i, j, k) = 2 \cdot k$ and $i + j > k$. So we can solve this problem in complexity $\mathcal O(n \sqrt n + \sum_{i = 1}^n \sigma_0(i))$ per testcase, which is faster than the editorial.code 168224359
•  » » 13 months ago, # ^ |   +8 Yes, but I don't really like number theory, so I didn't want to include more of it in the solution.
 » 13 months ago, # |   0 C feels amazing, I can only think of BFS to solve it
 » 13 months ago, # |   0 Very good contest, one of the best I have seen yet :)
 » 13 months ago, # | ← Rev. 2 →   0 personally liked all the anime references.
 » 13 months ago, # |   0 But I can't understand D. /sad
•  » » 13 months ago, # ^ | ← Rev. 2 →   0 Video Solution for D.(Video Solution for C.)
 » 13 months ago, # | ← Rev. 2 →   +3 Here is my proof for B. The key claim is the following:Claim: Let $a,b$ be positive integers such that $a\ne 1$ or $b\ne 1$ (or both), then $\operatorname{lcm}(a,b) \leq \frac{a^2+b^2-1}{2}$.Proof: If $a\neq b$, then $\displaystyle{\operatorname{lcm}(a,b)\leq ab = \frac{a^2+b^2-(a-b)^2}{2} \leq \frac{a^2+b^2-1}{2}}.$Otherwise, $a=b$, then $\displaystyle{\operatorname{lcm}(a,b) = a \leq \frac{2a^2-1}{2}}.$Hence, the claim is proven $\blacksquare$Now, we just use the claim (note the exception at $1$): $\displaystyle{\sum_{i=1}^n \operatorname{lcm}(i,a_i) \leq \frac 12 + \sum_{i=1}^n \frac{i^2+a_i^2-1}{2} = \sum_{i=1}^n i^2 - \frac{n-1}{2}}.$If $n$ is odd, then this is the desired bound. Otherwise, note that this quantity is not an integer, so we actually reduce the bound by $\tfrac 12$. One can check that this yields the equality.
•  » » 13 months ago, # ^ |   +8 this is very elegant!
 » 13 months ago, # |   +8 Bonus Problem's Solution for A We have to find $ans_1, ans_2, \ldots ans_n$ First Initialize each $ans_i$ to $0$. Then traverse the array p from end Now say element x appears at index i and $x  » 13 months ago, # | 0 About Bonus Problem of C If any$a_i$is$\le 0$and any$a_j$is$ \ge 0 $such that$j
 » 13 months ago, # |   +16 A-E2 video Editorial for Chinese ：Bilibili
 » 13 months ago, # |   0 Comeback after almost a year ... I did pretty bad
 » 13 months ago, # |   0 for E2. "Turns out that for the current constraints, for every k, we can iterate over all pairs of 1≤i
•  » » 13 months ago, # ^ |   +8 Judging by this answer, $\ll n \log^3 n$.
 » 13 months ago, # |   0 In editorial of problem D, what is the "another case to work"?
 » 13 months ago, # |   0 In problem E how can I find this $\frac{(r - l + 1) \cdot (r - l) \cdot (r - l - 1)}{6}$ .
•  » » 13 months ago, # ^ |   0 It's just $C_{r - l + 1}^3$ — the number of ways too choose three elements from $r - l + 1$ elements (if we don't care about the order).
 » 13 months ago, # |   0 What is the solution for F Bonus?
•  » » 13 months ago, # ^ |   0 Did you solve the original F problem first?
 » 13 months ago, # |   0 Can somebody help me out with Problem D : LinkMy approach is ans = max( min(2 * mn , min(a[i],a[i+1]))) , where mn is the minimum value of a.
 » 13 months ago, # |   0 Any ideas about Bonus Problem for D? I have solved problem D, but I can't think of any approach for doing it for all k.Maybe Vladithur can create a separate blog post about the solutions of bonus problems.
•  » » 13 months ago, # ^ |   0 I'm too lazy to do that(As for the bonus of problem D, the basic idea is to extend the greedy solution.
•  » » 13 months ago, # ^ |   0 If we try to extend the greedy solution of D. The complexity reaches O(n^2). To find for any k we need to do O(n) processing. I tried to think how to use the solution of k-1 to solve for k, But there seems a lot of casework, and even after that complexity still seems to be O(n^2) in some worst cases. Can someone help me here ?
•  » » » 13 months ago, # ^ | ← Rev. 3 →   0 What I meant is that you process the values of $a$ in increasing order and then maintain a couple of sets / multisets to find the answer for every $k$. Try to find what information you need to know to get the answer for some $k$ (and is easy to maintain when you move on to $k + 1$).
 » 13 months ago, # |   0 I found C to be bit on a tougher side, compared to the no of solves during the contest :/
 » 13 months ago, # |   0 I am very confused about the use of i+= i & (-i) loop from many of the fastest solutions of problem E2. For example, this, this and this.It seems to me that there are some hidden patterns that I am not aware of. Could anyone help me?
•  » » 13 months ago, # ^ |   0 It's just one possible implementation of a fenwick tree.
 » 13 months ago, # | ← Rev. 15 →   -6 .
 » 13 months ago, # |   0 Could any warm-hearted baby provide a previous code of D? I try my best but owing to my poor intelligence not to be able to do with it
•  » » 6 weeks ago, # ^ |   0 Joining 2 Hands Emoji
 » 13 months ago, # |   0 Can anyone tell me why I got the wrong answer in https://codeforces.com/contest/1712/submission/168392728 of problem d?
 » 13 months ago, # | ← Rev. 4 →   +8 Thanks for amazing round and tutorial but there is a small mistake Div2B Hint 3 lcm(x, x + 1) + lcm(x + 1, x) = x² + (x + 1)²-1
•  » » 13 months ago, # ^ |   0 Fixed
 » 13 months ago, # |   +10 why is this unrated now ?
 » 13 months ago, # |   0 We can probably prove problem B by induction? In addition, I don't understand this sentence in problem B: So we need to minimize the sum of $max(x1…xk)−min(x1…xk)$ over all cycles. Could anyone please explain it? Sory for my poor English.
 » 13 months ago, # |   0 "If there is at least one operation left, there are two cases: k=1 and k>=2.If the first case, it is optimal to apply our operation near one of the maximums in the array to maximize min(a_i,a_{i+1})"Would it not be more optimal to change a_i > = ans/2 to 1e9? Consider ans = 18, and the following array-[0,1,4,5,6,9,10,11,15,17] and nodes [1,2,3,4,5,6,7,8,9,10]Let k = 6. The binary search solution would turn this to-[1e9, 1e9, 1e9, 1e9, 1e9, 9, 10, 11, 15, 17] and k = 1.Now, if we change 15 to 1e9 (as mentioned in the solution), then d(9,10) = 2*9 = 18. But if we change 9 to 1e9, then d(6,8) = 2*10 = 20 > 18?
•  » » 13 months ago, # ^ |   0 I meant $k = 1$ and $k \ge 2$ for the value initially given, not after we apply the operation some number of times.
•  » » » 13 months ago, # ^ |   0 I am not sure I follow, consider-arr = [2,3,4,5,11,12], nodes = [1,2,3,4,5,6] and k = 1. Let ans = 5. In the binary search, no a_i's would be turned to 1e9 as 5/2 = 2 !< 2. First if condition fails as operation wasn't applied. In the second if condition, k==1 and max(a_1,..,a_n) = 12 > 5, hence it will return true.But, if we change 11 to 1e9, then d(5,6)= 2*min = 2*2 = 4 < 5?
•  » » » » 13 months ago, # ^ |   0 The division is not floored, $\frac{5}{2} = 2.5$, not $2$.
 » 13 months ago, # |   0 Why tutorial for problem E1 is missing?
•  » » 13 months ago, # ^ |   0 Because the main idea of the model solutions is the same in both versions, E2 just uses a data structure to speed up the last part of the solution, which you don't need to use in E1.
•  » » » 13 months ago, # ^ |   0 Thanks, it would be really helpful if you can mention this information at the start of the tutorial for E2. Or rather one should have made the editorial for E1 because I think people usually solve in order so they won't look up into E2 before they have solved E1 expecting that E2 must have a different solution.
 » 13 months ago, # |   0 Vladithur can u elaborate on the proof for greedy in div2 D. Why is it optimal to consider k-1 minimums in the first place?
•  » » 13 months ago, # ^ |   0 This is somewhat similar to the binary search solution. In it you change the smallest values such that all $a_i \ge \frac{ans}{2}$, and then cleverly choose which remaining values to change. Turns out that there are only two cases: zero values left or $\ge$ one value left, so in our greedy solution, we also leave one value to do something with, hence we change the $k - 1$ smallest values.
 » 13 months ago, # |   0 For Problem E2 : It is mentioned as Since ik) but there is no explanation like why it is true or how do the author arrive to this condition. Vladithur Can you explain this ? Like its is easy to observe that if lcm(i, j, k) = k then it would be bad, but for 2k I am not sure. Also why not the values in between k and 2k.
 » 13 months ago, # | ← Rev. 3 →   0 C :wHERE IS MISTAKE , iT'S FALLING ON 2ND TEST CASEhttps://codeforces.com/problemset/submission/1712/170962914
 » 12 months ago, # | ← Rev. 4 →   +8 Casework O(nlogn) greedy solution for D:Maintain array v and sorted array v2One possible solution is to reduce all the k minimum elements to 1e9 then ans=max(ans,min(v[i]+v[i+1],v2[k]*2)) ****as elements 0......k-1 in v2 assigned 1e9 in v so the smallest element in v is v2[k]Now reset vAnother possible solution is to assign the smallest k-1 elements to 1e9 then ans=max(ans,min(max(v[i],v[i+1]),v2[k-1]*2)); ****This is because we have an element v[i]. Next or previous element we can assign 1e9 so it becomes min(v[i],1e9) = max(v[i],v[i+1]) as we assign the minimum element to 1e9Now if k>=2 Another possible solution is to assign the smallest k-2 elements to 1e9 then ans=max(ans,min(1e9,v2[k-2]*2)) ****as we have two consecutive elements of the form [1e9 1e9] in the array and the smallest element in the array is v2[k-2]